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# Reasoning behind shifting over the text whem mismatch occurs in KMP algorithm?

I am not sure that you've got problems with understanding in only this point, so, if you don't mind, I'll just describe (with as much explanation as possible) the whole algorithm. The answer to your question is probably in the last paragraph, but you'd better read it all to understand my terminology better.

During the KMP algorithm, you are, actually, counting nearly the same values as in the table (this is usually called prefix function). So when you get to position i in the text, you need to count the maximum length of a substring in the text ending in position i which equals some prefix of the pattern. It is quite clear that if and only if the length of this substring is equal to length of the pattern, you have found the pattern in the text. So, how do you count this prefix function value fast? (I suppose that you count these for the pattern using some O(n^2) algorithm, which is not enough fast). Let's suppose that we've already did everything for the first `i-1` symbols of the text and we are now working with position `i`. We will also need the prefix-function value for the previous symbol of the text: `p[i-1]`.

Let's compare text[i] and pattern[p[i-1]] (indexation from 0, if you don't mind). We already know that ```pattern[0:p[i-1]-1] == text[i-1+p[i-1],i-1]```: that's the definition of `p[i-1]`. So, if ```text[i] == pattern[p[i-1]]```, we now know that `pattern[0:p[i-1]] == text[i-1+p[i-1]`, i]', and that's why p[i] = p[i - 1]. But the interesting part starts when ```text[i] != pattern[p[i-1]]```.

When these symbols are different, we start jumping. The reason for that is that we want to find the next possible prefix as fast as we can. So, how we do it. Just look at the picture here and follow the explanation (the yellow parts are substrings found for `text[i-1]`). We're trying to find some string `s`: `s:=s1+text[i]`. Because of the prefix-function definition, `s1=s2, c=test[i]`. But we already know (from finding the value for `text[i-1]`) that two yellow parts in the picture are the same, so `s3` actually equals `s1`, and so `s3=s1`. So we can find the length of `s1`: it is `table[p[i-1]]`. So now if `c1=text[i]`, we should stop: we've found `p[i]`, it is `s1.length + 1`. And if `c1!=text[i]`, we can just repeat the same jumping, looking now at the first `table[table[p[i-1]]]` symbols of the pattern, and so we go on until we find the answer, or we get to the first 0 symbols in this case `p[i]:=0`.

Categories : String

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